2
$\begingroup$

Consider the following two non-static situations.

  • When we connect a dc or ac source across a metal, an electric field exists inside the metal which causes the electrons to flow through the metal thereby constituting a current.

  • On the other hand, when an electromagnetic wave falls on a metal, the alternating electric field of the wave cannot penetrate deep into the metal. There is a very tiny skin depth.

Why does this difference arise?

$\endgroup$
2
  • 2
    $\begingroup$ The skin depth is proportional to sqrt(1/f). For DC currents the frequency is zero and the skin depth becomes infinite, i.e. the electric field can penetrate the metal to any depth. There is no difference because the skin depth formula can account for both cases. DC is simply the limit for f->0. Having said that, the skin depth is not all that deep. In copper it is roughly 9mm for 50Hz AC and it would be approx. 65mm for 1Hz currents. $\endgroup$ Oct 9, 2022 at 6:11
  • $\begingroup$ @FlatterMann Thanks! Can you write it as an answer? Can you explain this physically? $\endgroup$ Oct 9, 2022 at 6:14

1 Answer 1

5
$\begingroup$

There is no difference because the skin depth formula can account for both cases. The skin depth is proportional to ${\sqrt\frac{1}{f}}$. For DC currents we can take the limit of the frequency $\lim f\to0$ and the skin depth becomes infinite, i.e. the electric field can penetrate the metal to any depth. Having said that, the skin depth is not all that deep. In copper it is roughly 9mm for 50Hz AC and it would be approx. 65mm for 1Hz currents.

$\endgroup$
5
  • $\begingroup$ Depending on your perspective, 9mm might be very deep indeed. I suppose the "not all that deep" characterization imagines an EM wave falling on the end of a wire, but what about one falling on a foil, or on a piece of sheet metal, or on the traces on a circuit board? $\endgroup$ Oct 9, 2022 at 16:08
  • $\begingroup$ This explains the DC case, but leaves open how wires can transmit AC currents. (To be precise: when we expose the two ends of a wire to AC voltage sources, why do we observe AC current throughout the wire, as opposed to just the 9mm strips around the ends?) $\endgroup$ Oct 9, 2022 at 16:38
  • $\begingroup$ @JohnBollinger You are correct, if one wants to use copper as shielding for magnetic fields at 50/60Hz, then 9mm is incredibly deep and usually too heave and too expensive. The folks who are building GW scale AC transmission lines, however, will probably see it as "shallow". At the scale (and metal thickness) of circuit boards the dividing line is somewhere in the tens of kHz to low MHz range. I managed to suppress magnetic noise from a DC/DC converter with several layers of copper tape once. It was an easy fix. So, yeah, it depends. $\endgroup$ Oct 9, 2022 at 20:49
  • 1
    $\begingroup$ @JacobManaker The electromagnetic energy flow is always in the electromagnetic field. The "Poynting vector" (the cross product between magnetic field and electrical field) is the relevant physical quantity that describes energy flow. What the high density of electrons inside a wire does is to form boundary conditions for the EM field that contain the majority of the energy density inside and near the wire. At DC the electric field can penetrate the wire, so there is a non-zero Poynting vector inside, but at AC less of the field can penetrate and most of the energy flows on the outside. $\endgroup$ Oct 9, 2022 at 20:57
  • 1
    $\begingroup$ @JacobManaker In short, the energy does not flow through the wire in the first place. But I won't attempt an explanation as Veritasium has two videos that explain the answer to your question beautifully: youtube.com/watch?v=bHIhgxav9LY youtube.com/watch?v=oI_X2cMHNe0 The first of the two is the easier to understand, while the second clears up all the gory details where the first video rounded off some corners. Both are definitely worth a watch, imho. $\endgroup$ Oct 9, 2022 at 21:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.